If the value of the integral $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{(\sin x)^{2023}}}\right) dx = \frac{\pi}{4}(\pi + a) - 2$$, then the value of $$a$$ is

We need to evaluate $$\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\right)dx$$.

Evaluate $$I_1 = \int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+\pi^x}\,dx$$.

We use the property: for even function $$f(x)$$ and constant $$b > 0$$,

$$\int_{-a}^{a}\frac{f(x)}{1+b^x}\,dx = \int_0^a f(x)\,dx$$

Here $$f(x) = x^2\cos x$$ is even (product of two even functions) and $$b = \pi$$.

$$I_1 = \int_0^{\pi/2} x^2\cos x\,dx$$

Using integration by parts twice:

$$\int x^2\cos x\,dx = x^2\sin x - 2\int x\sin x\,dx = x^2\sin x - 2\left[-x\cos x + \sin x\right]$$

$$= x^2\sin x + 2x\cos x - 2\sin x$$

Evaluating from $$0$$ to $$\frac{\pi}{2}$$:

$$I_1 = \left[\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right] - [0] = \frac{\pi^2}{4} - 2$$

Evaluate $$I_2 = \int_{-\pi/2}^{\pi/2}\frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\,dx$$.

Let $$g(x) = (\sin x)^{2023}$$. Since 2023 is odd, $$g(-x) = (-\sin x)^{2023} = -g(x)$$, so $$g$$ is an odd function.

The function $$h(x) = 1 + \sin^2 x$$ is even.

Using the same property (with $$e^{g(x)}$$ playing the role of $$b^x$$ where $$g$$ is odd):

$$I_2 = \int_0^{\pi/2}(1+\sin^2 x)\,dx$$

$$= \int_0^{\pi/2} 1\,dx + \int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$$

(Using $$\int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{4}$$.)

Combine and compare.

$$I_1 + I_2 = \frac{\pi^2}{4} - 2 + \frac{3\pi}{4} = \frac{\pi}{4}(\pi + 3) - 2$$

Comparing with $$\frac{\pi}{4}(\pi + a) - 2$$, we get:

$$a = 3$$

The correct answer is Option (1): $$\boxed{3}$$.